For second-order differental equations with linear constant coefficients, we can try the solution \(y = e^{kt}\):
\( ay^{''} + by^{'} + cy = 0 \)
\( ak^2e^{kt} + bke^{kt} + ce^{kt} = 0 \)
\( e^{kt}(ak^2 + bk + c) = 0 \)
The exponential term is never zero, thus the quadratic term with respect to \(k\) needs to be 0. This equation is the characteristic equation or auxilliary equation:
\(ak^2 + bk + c = 0\)
Recall that the quanity \(b^2 - 4ac\) is the determinant. We can solve for values of \(k\) that make this characteristic equation true. There are three possible outcomes:
| Scenario | Determinant | Solution |
| 2 Real Roots | \(b^2 - 4ac > 0\) | \( y = c_1 e^{k_1t} + c_2 e^{k_2t} \) |
| Repeated Real Roots | \(b^2 - 4ac = 0\) | \( y = c_1 e^{k_1t} + c_2 te^{k_1t} \) |
| Complex Roots | \(b^2 - 4ac < 0\) | \( y = e^{\alpha t}(c_1 \cos {\beta t} + c_2 \sin {\beta t})\) |
Solve the DE \( y^{''} - 16y^{'} + 48y = 0 \)
This is a second-order linear DE with constant coefficients. We can try \( y = e^{kt}\) and solve the characteristic equation:
\( k^2 - 16k + 48 = 0 \)
\( b^2 - 4ac = (-16)^2 - 4(1)(48) \)
\( b^2 - 4ac = 64 > 0\)
The determinant is positive meaning there are two real roots. Factoring gives:
\((k-4)(k-12) = 0 \)
The general solution is \(\boldsymbol{y = c_1 e^{4t} + c_2 e^{12t}}\).
Solve the DE \( y^{''} - 10 y^{'} + 25y = 0 \)
Again, this is a second-order linear DE with constant coefficients. Calcualte the determinant:
\( b^2 - 4ac = (10)^2 - 4(1)(25) = 0 \)
The determinant is zero meaning there are repeated roots. Factoring gives:
\((k-5)(k-5) = 0 \)
The general solution is \(\boldsymbol{y = c_1 e^{5t} + c_2 te^{5t}}\).
Note that the second term is actually found using the reduction of order technique.
Solve the DE \( y^{''} + 25y = 0 \)
Again, this is a second-order linear DE with constant coefficients. Calcualte the determinant:
\( b^2 - 4ac = (0)^2 - 4(1)(25) = -100 < 0 \)
The determinant is negative meaning there are complex roots. Using the quadratic equation gives:
\( \cfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \cfrac{-0 \pm (10i)}{2} = 0 \pm 5i \)
The general solution is \( y = c_1 \cos {5 t} + c_2 \sin {5 t} \) since \(\alpha = 0\).
Note that this technique expands to higher-order homogeneous differential equations with constant coefficients. Suppose the characteristic equation is:
\( (k+2)^2(k-1)(k^2-2k+3) \)
The first term has repeated roots giving solution \(c_1e^{-2t} + c_2 t e^{-2t}\).
The second term has a real root giving solution \(c_3e^t\).
The third term has complex roots in the form \( 1 \pm \sqrt{2} \) giving solution \( e^{ t}(c_4 \cos {\sqrt{2} t} + c_5 \sin {\sqrt{2} t}) \)
Putting it all together gives the general solution to the DE:
\( y = c_1e^{-2t} + c_2 t e^{-2t} + c_3e^t + e^{ t}(c_4 \cos {\sqrt{2} t} + c_5 \sin {\sqrt{2} t})\).